1 / 11

6.1 Integration by parts

6.1 Integration by parts. Formula for Integration by parts. The idea is to use the above formula to simplify an integration task. One wants to find a representation for the function to be integrated in the form udv so that the function vdu is easier to integrate

oya
Download Presentation

6.1 Integration by parts

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 6.1 Integration by parts

  2. Formula for Integration by parts • The idea is to use the above formula to simplify an integration task. • One wants to find a representation for the function to be integrated • in the form udvso that the function vdu is easier to integrate • than the original function. • The rule is proved using the Product Rule for differentiation.

  3. Deriving the Formula Start with the product rule: This is the Integration by Parts formula.

  4. Logs, Inverse trig, Polynomial, Exponential, Trig Choosinguandv dv is easy to integrate. u differentiates to zero (usually). Choose u in this order: LIPET

  5. Example 1: LIPET polynomial factor

  6. Example 2: LIPET logarithmic factor

  7. Example 3: LIPET This is still a product, so we need to use integration by parts again.

  8. Example 4: LIPET This is the expression we started with!

  9. Example 4(cont.): LIPET This is called “solving for the unknown integral.” It works when both factors integrate and differentiate forever.

  10. Integration by Parts for Definite Integrals Formula Integration by Parts Formula and the Fundamental Theorem of Calculus imply the above Integration by Parts Formula for Definite Integrals. Here we must assume that the functions u and v and their derivatives are all continuous. Example

  11. Integration by Parts for Definite Integrals Example (cont’d) By the computations on the previous slide we now have Combining these results we get the answer

More Related